Quantitative Aptitude PYQs

Concept:

Total number of factors of x^py^q is (p + 1)(q + 1)

Number of odd factors of x^py^qz^r if x is even and y and z are odd then the factors are

1× (q + 1)(r + 1)

Calculation:

We have, 

⇒ N = 12⁶ × 3⁸ × 5³

⇒ N = 2⁶ × 2⁶ × 3⁶ × 3⁸ × 5³

⇒ N = 2¹² × 3¹⁴ × 5³

Total number of factors:- (12 + 1) × (14 + 1) × (3 + 1)

Total number of factors:- 13 × 15 × 4

Total number of factors = 780

Statement I: The number of odd factors of N is 60.

Number of odd factors = 1 × (14 + 1) × (3 +1)

Number of odd factors = 1 × 15 × 4

A number of odd factors = 60.

Hence, Statement I is correct.

Statement II: The number of even factors of N is 720.

Number of even factors = Total number of factors - Odd factors

Number of even factors = 780 - 60

Number of even factors = 720

Hence, Statement II is correct.

∴ Both statements are correct.

Given:

 \(\rm \frac{\log_{10}a}{b-c}=\frac{\log_{10}b}{c-a}=\frac{\log_{10}c}{a-b}\) and (a ≠ b ≠ c) 

Formula Used:

log(m × n) = log m + log n

Calculation:

We have,

⇒  \(\rm \frac{\log_{10}a}{b-c}=\frac{\log_{10}b}{c-a}=\frac{\log_{10}c}{a-b}\) = k (say)

⇒ log₁₀a = k(b - c)      ------ (1)

⇒ log₁₀b = k(c - a)      ----- (2)

⇒ log₁₀c = k(a - b)     ------ (3)

Adding eq. (1), (2) and (3)

⇒ log₁₀a + log₁₀b + log₁₀c = k(b - c + c - a + a - b)

⇒ log₁₀a + log₁₀b + log₁₀c = k × 0

⇒ log₁₀a + log₁₀b + log₁₀c = 0

⇒ log₁₀(abc) = 0

⇒ abc = 10⁰

⇒ abc = 1

∴ If \(\rm \frac{\log_{10}a}{b-c}=\frac{\log_{10}b}{c-a}=\frac{\log_{10}c}{a-b}\), (a ≠ b ≠ c), then the value of abc is 1.

Formula used:

Angle between the hands = 11 × \(\frac{m}{{2}}\) − 30H

Where, m denotes the minute hand and H denotes the hours hand

Calculation:

According to the question

Angle between the hands = 11 × \(\frac{m}{{2}}\) − 30H

⇒ (11 × \(\frac{40}{{2}}\) − 30 × 4)

⇒ (11 × 20 − 120)

⇒ (220 − 120)

⇒ 100° 

∴ The required angle is 100°.

Given:

log10\(\rm \left[995+\sqrt{x^2-12x+60}\right]=3\)

Concept:

Quadratic Equation:- A quadratic equation is a second order equation written as ax² + bx + c = 0 where a, b and c are coefficients of real numbers and a ≠ 0.

If α and β are the roots of the quadratic equation

Sum of roots;- α + β = \(\frac{-b}{a}\)

Formula Used:

log_ax^m = mlog_ax

Calculation:

We have,

⇒ log10\(\rm \left[995+\sqrt{x^2-12x+60}\right]=3\)

⇒ \(log_{10} \left[ 995 \ + \ \sqrt{x^2 \ - \ 12x \ + \ 60} \right] = log_{10}10^3\)

⇒ \(995 \ + \ \sqrt{x^2 \ - \ 12x \ + 60} = 10^3\)

⇒ \(995 + \sqrt{x^2 \ - 12x \ + \ 60} = 1000\)

⇒ \(\sqrt{x^2 \ - \ 12x \ + 60} = 5\)

On squaring both the sides, we get

⇒ x² - 12x + 60 = 25

⇒ x² - 12x + 35 = 0

⇒ Sum of roots = \(\frac{-(-12)}{1}\)= 12

∴ If log10\(\rm \left[995+\sqrt{x^2-12x+60}\right]=3\), then the sum of the roots of the equation is 12.

Formula used:

1. For quadratic equation ax² + bx + c = 0

\(x = {-b \pm \sqrt{(b^2-4ac)} \over 2a}\) 

2. (a + b)(a - b) = a2 - b2

3. (a + b)² = a² + b² + 2ab

Calculation:

qx2 – 2px + q = 0 

By above formula,

\(x = {-b \pm \sqrt{(b^2-4ac)} \over 2a}\)

Here, a = q, b = -2p and c = q

\(x = {2p \pm \sqrt{(4p^2-4q^2)} \over 2q}\)

⇒ \(x = {p \pm \sqrt{p^2-q^2} \over q}\)     ----(1)

Statement:

1 

x = \(\rm \frac {\sqrt{p + q}+\sqrt{p-q}}{\sqrt{p+q}-\sqrt{p-q}}\), where p > q

Rationalizing the denominator

⇒ \(\rm \frac {\sqrt{p + q}+\sqrt{p-q}}{\sqrt{p+q}-\sqrt{p-q}}\) × \(\rm \frac {\sqrt{p + q}+\sqrt{p-q}}{\sqrt{p+q}+\sqrt{p-q}}\)

Using the identities (2) & (3)

⇒ \(x = \frac{p+q+p-q+2\sqrt{(p+q)(p-q)}}{p+q+p-q}\)

⇒ \(x = \frac{p+q+p-q+2\sqrt{(p^2-q^2)}}{p+q+p-q}\)

⇒ \(x = {p \pm \sqrt{p^2-q^2} \over q}\)

Hence, statement 1 is correct

Statement: 2 

From equation (2), we can say that,

\(x = {p + \sqrt{p^2-q^2} \over q}\)

Hence, statement 2 is also correct.

Statement 3: 

We have, \(x = {p \pm \sqrt{p^2-q^2} \over q}\)

For any value of p & q, 

x = \(\rm \frac {\sqrt{p + q}}{\sqrt{p-q}}\) is not possible

Hence, statement 3 is wrong

∴ Only two values of x are possible.

Shortcut Trick

Use the standard polynomial identity: LCM × GCD = p(x) × q(x).

Given p(x) = x⁴ + x² + 1 and GCD = x² − x + 1.

Note that x⁴ + x² + 1 = (x² + x + 1)(x² − x + 1).

LCM = [p(x) × q(x)] ÷ GCD = [(x² + x + 1)(x² − x + 1) × q(x)] ÷ (x² − x + 1).

LCM = (x² + x + 1) × q(x). Since q(x) = (x² − x + 1)², LCM = (x² + x + 1)(x² − x + 1)².

∴ The correct answer is (x² + x + 1)(x² − x + 1)².

Alternate Method

Given:

p(x) = x⁴ + x² + 1

q(x) = x⁴ − 2x³ + 3x² − 2x + 1

GCD = x² − x + 1

Formula Used:

LCM(p, q) = [p(x) × q(x)] ÷ GCD(p, q)

Calculations:

⇒ Factorizing p(x): x⁴ + x² + 1 = (x⁴ + 2x² + 1) − x² = (x² + 1)² − x²

⇒ p(x) = (x² + 1 + x)(x² + 1 − x) = (x² + x + 1)(x² − x + 1)

⇒ Factorizing q(x): x⁴ − 2x³ + 3x² − 2x + 1 = (x² − x + 1)²

⇒ Verification: (x² − x + 1)(x² − x + 1) = x²(x² − x + 1) − x(x² − x + 1) + 1(x² − x + 1)

⇒ x⁴ − x³ + x² − x³ + x² − x + x² − x + 1 = x⁴ − 2x³ + 3x² − 2x + 1 (Correct)

⇒ LCM = [(x² + x + 1)(x² − x + 1) × (x² − x + 1)²] ÷ (x² − x + 1)

⇒ LCM = (x² + x + 1) × (x² − x + 1)²

∴ The correct answer is (x² + x + 1)(x² − x + 1)².

Additional Information

Polynomial Product Rule

For any two polynomials p(x) and q(x), the product of the polynomials is equal to the product of their LCM and GCD.

Sophie Germain Identity

The factorization x⁴ + x² + 1 = (x² + x + 1)(x² − x + 1) is a specific case of the Sophie Germain Identity used frequently in algebra.

GCD Definition

The Greatest Common Divisor (GCD) is the highest degree polynomial that divides both p(x) and q(x) without a remainder.

Calculation:

According to the question

First 2 digit number = 10x + y

By reversing the 2-digit number = 10y + x

Adding the two numbers we get,

⇒ (10x + y) + (10y + x)

⇒ 11(x + y)

Statement:1 The sum is divisible by 9

Put x = 9, y = 9

⇒ 11 (9 + 9) = 198

Which is divisible by 9

Statement: 2 The sum is divisible by 2

Put x = 2, y = 2

⇒ 11 (2 + 2) = 44

Which is divisible by 2

∴ Both Statement-1 and Statement-2 are sufficient to answer

the question

Given:

LCM of two numbers x and y is 481 where x > y.

Concept used:

HCF: Product of the smallest power of each common prime factor in the numbers.

LCM: Product of the greatest power of each prime factor, involved in the numbers

Calculation:

The factors of 481 = (481 × 1) and (37 × 13)

According to the question x > y, then 

(x, y) = (481, 1) or (37, 13)

Statement 1:  y > 1

Hence, only possible value are

x = 37, y = 13

We can easily find the value of (3x – 2y).

Statement 2. HCF of x and y is 1

In both (481, 1) or (37, 13), HCF is 1.

Two different values of (3x – 2y)

∴ Statement-1 alone is sufficient to answer the question

Concept used:

Natural number: All the numbers start from 1 till infinity.

The factor of  xⁿ + yⁿ: 

When n is odd ⇒ (x + y)

When n is even ⇒ No factor

Explanation:

Statement-1: n is a natural number

When n is odd, a factor of (xn + yn) is (x + y) but,

No factor in the case when n is even

Therefore, with this statement, we are not sure about the factor of (x + y). For the odd value of n, (x + y) will be a factor & for the even value of n, (x + y) will not be a factor.

Statement-2: n is an even natural number

We have discussed already when n is even, there is no factor.

In this case (xn + yn) will not be divisible by (x + y)

Hence, if we consider only statement 2 (i.e. n = even), we can surely say (x + y) is not divisible by (xⁿ + yⁿ). 

∴ Statement-2 alone is sufficient to answer the question.

Important PointsThe question says, which one of the following is correct in respect of the question and the statements?

It doesn't mean that we have to determine whether statements 1 or 2 is/or correct or not. 

We have to find the correct option using two statements. Clearly, with statement 2, we are sure about the answer i.e. "if n = even, (x + y) will not be a factor of (xⁿ + yⁿ)'. So it is sufficient to answer.

Given:

Let x and y be two real numbers. 

Concept:  

1. For positive real number:

• Both even & odd power gives positive result.
• Ex: (2)² = 4, (-2)² = 4
  2. For negative real number:
• Even power gives positive result
• Odd power gives negative result.
• Ex: (2)³ = 8, (-2)³ = -8

 

Calculation:

Statement 1:

x8y9 < 0

As x⁸ will be always positive 

⇒ y⁹ < 0    (product of +ve and -ve is negative) 

⇒ y < 0

Statement: 2

x9y10 < 0

As y¹⁰ will be always positive 

⇒ x9 < 0    (product of +ve and -ve is negative) 

⇒ x < 0

From 1 & 2, we can say x < 0 & y < 0

Since, product of -ve and -ve is positive

⇒ xy > 0

∴ Both Statement-1 and Statement-2 are sufficient to answer the question

Given:

a = 3b, 4b = 5c, 6c = 7d

Calculation:

a = 3b, 4b = 5c, 6c = 7d

⇒ \(\frac{a}{{b}}\) = \(\frac{3}{{1}}\), \(\frac{b}{{c}}\) = \(\frac{5}{{4}}\), \(\frac{c}{{d}}\) = \(\frac{7}{{6}}\)

⇒ (\(\frac{a}{{b}}\) × \(\frac{b}{{c}}\) × \(\frac{c}{{d}}\)) = (\(\frac{3}{{1}}\) × \(\frac{5}{{4}}\) × \(\frac{7}{{6}}\))

⇒ \(\frac{a}{{d}}\) = \(\frac{35}{{8}}\)

Now, the value of \(\rm\frac{d+a}{d-a}\)

⇒ \((\rm\frac{8+35}{8-35})\) = \(-\rm\frac{43}{27}\)

∴ The required value is \(-\rm\frac{43}{27}\).

Given:

The given polynomial = (k - 3) x2 - kx - 1

Concept used:

If ax² + bx + c = 0 has no real roots, then

Discriminant (D) = b²- 4ac < 0

Calculation:

Given quadratic equation is (k - 3) x2 - kx - 1

According to the above concept,

⇒ b² - 4ac < 0

⇒ (-k)² - 4(k - 3)(-1) < 0

⇒ k² + 4k - 12 < 0

⇒ k² + 6k - 2k - 12 < 0

⇒ k(k + 6) - 2(k + 6) < 0 

⇒ (k - 2)(k + 6) < 0

∴ The required value is -6 < k < 2.

Given:

bc + cd = 2bd

a + c = 2b

Shortcut Trick

Put a = 1, b = 2 or c = 3

⇒ bc + cd = 2bd

⇒ (2 × 3) + 3d = 2 × 2 × d

⇒ 6 + 3d = 4d

⇒ d = 6

From option 3:

ad − bc = 1 × 6 - 2 × 3 = 0

∴ Option 3 is right.

Alternate Method

According to the question

bc + cd = 2bd     -----(1)

a + c = 2b

Putting the value of 2b in equation (1) then,

⇒ bc + cd = (a + c) × d

⇒ bc + cd = ad + cd

⇒ bc = ad

⇒ ad − bc = 0

∴ The required value is ad − bc = 0

Given:

Distance covered by flight = 2800 km

Average speed reduced by 100 km/hr

Time of flight increased by 30 min = 1/2 hour

Formula Used:

Distance = Speed × Time

Shortcut Trick

We have,

⇒ \(\frac{Distance}{Speed} = Time\)

⇒ \(\frac{2800}{(x - 100)} - \frac{2800}{x} = \frac{1}{2}\)

Put x = 800

⇒ \(\frac{2800}{700} - \frac{2800}{800} = time\)

⇒ \(4 - \frac{7}{2} = time\)

⇒ 0.5 = \(\frac{1}{2} \) = time

∴ The original average speed of the aircraft is 800 km/hr.

Alternate MethodLet the original average speed (S₁) = x

Average speed reduced by 100(S₂)  = x - 100

Total speed ΔS = S₂ - S₁ = x - x - 100 = 100

⇒ D = \(\frac{S_1 \ \times S_2}{\Delta S} \times \Delta t\)

⇒ \(2800 = \frac{x(x - 100)}{100} \times \frac{1}{2}\)

⇒ 560000 = x(x - 100)

⇒ x² - 100x - 560000 = 0

⇒ x² -  800x + 700x - 560000 = 0

⇒ x(x - 800) + 700(x - 800) = 0 

⇒ (x - 800)(x + 700) = 0

⇒ x = 800 or x = - 700

∴ The original average speed of the aircraft is 800 km/hr.

Shortcut TrickPut a = 1, b = 0, c = 0

x = b + c = 0

y = c + a  = 1

z = a + b = 1

⇒ (x + y + z)3 – 24xyz = (0 + 1 + 1)³ - 24 × 0

⇒ (x + y + z)3 – 24xyz = 8

From option (3)

8 (a3 + b3 + c3) = 8(1³ + 0 + 0) = 1

∴ 8 (a3 + b3 + c3) is the correct answer.

Alternate Method

Formula Used:

(a + b + c)³ = a³ + b³ + c³ + 3(a + b)(b + c)(c + a)

Calculation:

We have,

⇒ (x + y + z)3 – 24xyz

⇒ (b + c + c + a + a + b)³ - 24(b + c)(c + a)(a + b)

⇒ [2(a + b + c)]³ - 24(b + c)(c + a)(a + b)

⇒ 8(a + b + c)³ - 24(b + c)(c + a)(a + b)

⇒ 8[a³ + b³ + c³ + 3(a + b)(b + c)(c + a)] - 24(b + c)(c + a)(a + b)

⇒ 8(a³ + b³ + c³) + 24(a + b)(b + c)(c + a) - 24(b + c)(c + a)(a + b)

⇒ 8(a³ + b³ + c³)

∴ The value of (x + y + z)3 – 24xyz is equal to 8(a³ + b³ + c³).

Given:

A person X and another person Y walk in the same direction

Average speed of X = u km/hr

Average speed of Y = v km/hr

Concept Used:

When two objects move in the same direction, then

Relative speed = S1 - S2

Formula used:

Speed = Distance/Time

Calculation

Time = Distance/Speed

⇒ t = d/(u − v)

Distance covered by X to overtake Y,

⇒ d = [(u × d/(u − v)]

⇒ d = \(\rm\frac{ud}{(u-v)}\)

∴ The required value is  \(\rm\frac{ud}{(u-v)}\).

Shortcut TrickLet the number be 2

Since p is the difference between a number and its reciprocal 

⇒ 2 - 1/2 = 3/2 = p

⇒ p⁴ + 4p² = p²(p² + 4) = (9/4)[9/4 + 4]

⇒ p4 + 4p2 = 225/16

Since q is the difference between the square of the same number

and the square of its reciprocal

⇒ (2)² - (1/2)² = 15/4 = q

⇒ q² = 225/16

∴ The required value is q2

Alternate Method

Formula used:

1. (a² − b²) = (a − b)(a + b)
2. (a − b)² = a² + b² − 2ab
3. (a + b)² = a² + b² + 2ab

 

Calculation:

Let the number be n

⇒ n − \(\frac{1}{{n}}\) = p          ------(1)

⇒  n² − \(\frac{1}{{n^2}}\) = q       -----(2)

⇒ q = (n − \(\frac{1}{{n}}\)) (n + \(\frac{1}{{n}}\))    [Using formula (1)]

⇒ q = p × (n + \(\frac{1}{{n}}\))            [From equation (1)]

⇒ q/p = (n + \(\frac{1}{{n}}\))             ------(3)

The value of  p4 + 4p²

⇒ p²(p² + 4)

⇒ p²(n − \(\frac{1}{{n}}\))² + 4

⇒ p²(n² + \(\frac{1}{{n^2}}\) − 2 + 4)       [Using formula (2)]

⇒ p²(n² + \(\frac{1}{{n^2}}\)+ 2)

⇒ p²(n + \(\frac{1}{{n^2}}\))²                        [Using formula (3)]

⇒ p² × \(\frac{q^2}{{p^2}}\) = q²                    [From equation (3)]

∴ The required value is q²

Given:

x = 2 + 2^1/2

Formula Used:

(a - b)² = a² + b² - 2ab

Calculation:

We have,

⇒ x = 2 + 2^1/2

⇒ x - 2 = 2^1/2

On squaring both the sides, we get

⇒ (x - 2)² = (2^1/2)²

⇒ (x - 2)² = 2

⇒ x² - 4x + 4 = 2

⇒ x² - 4x + 4 - 2 = 0

⇒ x² - 4x + 2 = 0

⇒ x² + 2 = 4x

On dividing by 'x', we get

⇒ \(x \ + \ \frac{2}{x} \ = \ 4\)

Again squaring on both the sides, we get

⇒ \(x^2 \ + \ \frac{4}{x^2}\ + \ 4 = 16\)

⇒ \(x^2 \ + \ \frac{4}{x^2} = 12\)

Again squaring on both the sides, we get

⇒ \(x^4 \ + \ \frac{16}{x^4} \ +\ 8 = 144\)

⇒ \(x^4 \ + \ \frac{16}{x^4 } = 136\)

∴ The value of x4 + 16x-4 is 136.

Given:

Time is taken by X longer than Y = 3 Hours

Distance = 30 km

Time taken by X if doubles his speed = 2 hours is less than Y 

**Formula used:**​

Let the speed of Y be u km/h & the speed of X be v km/h

According to the question

T_x = T_y + 3

30/u = 30/v + 3         ----(1)

Now, If X doubles his speed

30/2u = 30/v - 2        -----(2)

Equation (1) - (2)

(30/u - 30/2u) = (30/v + 3) - (30/v - 2)

30/2u = 5 or u  = 3 

Therefore, from equation (1)

30/3 = 30/v + 3

30/v = 7 or v = 30/7 = \(4\frac{2}{7}\) km/hr

∴ The required value of Y is \(4\frac{2}{7}\) km/hr.

Shortcut Trick

1% S.I interest in 10000 for 1^st quarter = 100 Rs

4% S.I interest in 10000 1^st quarter = 400 Rs

We know that, after the first quarter, C.I > S.I

∴ C > Rs. 400

Alternate Method

Formula used:

CI = [P(1 + R/100)T − P]

Where, P = Principal, R = Rate of interest and T = Time 

Calculation:

According to the question

Principal = Rs. 10,000

Rate = 4% per annum quarterly = (4/4)% = 1%

Time = 4 year  [1year = 4 quarters]

CI = [P(1 + R/100)T − P]

⇒ Rs. [10,000(1 + 1/100)⁴ − 10,000]

⇒ Rs. [10,000(1 + 0.01)⁴ − 10,000]

⇒ Rs. [10,000(1.01)⁴ − 10,000]

⇒ Rs. (10,000 × 1.0406 − 10,000)

⇒ Rs. (10,406 − 10,000)

⇒ Rs. 406

∴ C > Rs. 400